Lemma 53.19.17. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme equidimensional of dimension $1$ whose singularities are at-worst-nodal. If $Y \subset X$ is a reduced closed subscheme equidimensional of dimension $1$, then

the singularities of $Y$ are at-worst-nodal, and

if $Z \subset X$ is the scheme theoretic closure of $X \setminus Y$, then

the scheme theoretic intersection $Y \cap Z$ is the disjoint union of spectra of finite separable extensions of $k$,

each point of $Y \cap Z$ is a node of $X$, and

$Y \to \mathop{\mathrm{Spec}}(k)$ is smooth at every point of $Y \cap Z$.

**Proof.**
Since $X$ and $Y$ are reduced and equidimensional of dimension $1$, we see that $Y$ is the scheme theoretic union of a subset of the irreducible components of $X$ (in a reduced ring $(0)$ is the intersection of the minimal primes). Let $y \in Y$ be a closed point. If $y$ is in the smooth locus of $X \to \mathop{\mathrm{Spec}}(k)$, then $y$ is on a unique irreducible component of $X$ and we see that $Y$ and $X$ agree in an open neighbourhood of $y$. Hence $Y \to \mathop{\mathrm{Spec}}(k)$ is smooth at $y$. If $y$ is a node of $X$ but still lies on a unique irreducible component of $X$, then $y$ is a node on $Y$ by the same argument. Suppose that $y$ lies on more than $1$ irreducible component of $X$. Since the number of geometric branches of $X$ at $y$ is $2$ by Lemma 53.19.7, there can be at most $2$ irreducible components passing through $y$ by Properties, Lemma 28.15.5. If $Y$ contains both of these, then again $Y = X$ in an open neighbourhood of $y$ and $y$ is a node of $Y$. Finally, assume $Y$ contains only one of the irreducible components. After replacing $X$ by an open neighbourhood of $x$ we may assume $Y$ is one of the two irreducble components and $Z$ is the other. By Properties, Lemma 28.15.5 again we see that $X$ has two branches at $y$, i.e., the local ring $\mathcal{O}_{X, y}$ has two branches and that these branches come from $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{Z, y}$. Write $\mathcal{O}_{X, y}^\wedge \cong \kappa (y)[[u, v]]/(uv)$ as in Remark 53.19.9. The field $\kappa (y)$ is finite separable over $k$ by Lemma 53.19.7 for example. Thus, after possibly switching the roles of $u$ and $v$, the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa (y)[[u, v]]/(uv) \to \kappa (y)[[u]]$ and the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa (y)[[u, v]]/(uv) \to \kappa (y)[[v]]$. The scheme theoretic intersection of $Y \cap Z$ is cut out by the sum of their ideas which in the completion is $(u, v)$, i.e., the maximal ideal. Thus (2)(a) and (2)(b) are clear. Finally, (2)(c) holds: the completion of $\mathcal{O}_{Y, y}$ is regular, hence $\mathcal{O}_{Y, y}$ is regular (More on Algebra, Lemma 15.43.4) and $\kappa (y)/k$ is separable, hence smoothness in an open neighbourhood by Algebra, Lemma 10.140.5.
$\square$

## Comments (0)